$\textit{Đáp án + Giải thích các bước giải:}$
$\text{A = $\dfrac{3\sqrt{x} + 14}{\sqrt{x} + 2}$ ($x^{}$ $\geq$ 0)}$
$\text{A = $\dfrac{3\sqrt{x} + 6 + 8}{\sqrt{x} + 2}$}$
$\text{A = $\dfrac{3(\sqrt{x} + 2) + 8}{\sqrt{x} + 2}$}$
$\text{A = $\dfrac{3(\sqrt{x + 2})}{\sqrt{x} + 2}$ + $\dfrac{8}{\sqrt{x} + 2}$}$
$\text{A = 3 + $\dfrac{8}{\sqrt{x} + 2}$}$
$\text{Có $\sqrt{x}$ $\geq$ 0 $\forall$ $x^{}$}$
$\text{⇒ $\sqrt{x}$ + 2 $\geq$ 2}$
$\text{⇒ $\dfrac{1}{\sqrt{x} + 2}$ $\leq$ $\dfrac{1}{2}$}$
$\text{⇒ 3 + $\dfrac{8}{\sqrt{x} + 2}$ $\leq$ 7}$
$\text{Dấu "=" xảy ra khi $\dfrac{8}{\sqrt{x} + 2}$ = 4 = $\dfrac{8}{2}$}$
$\text{⇒ $\sqrt{x}$ + 2 = 2}$
$\text{⇒ $\sqrt{x}$ = 2 - 2 = 0}$
$\text{⇒ $x^{}$ = 0}$
$\text{Vậy GTLN của A = 7 khi $x^{}$ = 0}$
